Prove that $$S_n=\frac1{1^2}+\frac1{2^2}+\ldots+\frac1{n^2}<2$$for every $n\in\mathbb N$.
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Tags: Sequences, inequalities
22.06.2021 04:53
jasperE3 wrote: Prove that $$S_n=\frac1{1^2}+\frac1{2^2}+\ldots+\frac1{n^2}<2$$for every $n\in\mathbb N$. Use Induction $\frac1{1^2}+\frac1{2^2}+\ldots+\frac1{n^2}\leq 2-\frac{1}{n}$
22.06.2021 05:01
This is not really a prove lol: We have $$\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}<2,$$and that finishes
22.06.2021 05:04
math31415926535 wrote: This is not really a prove lol: We have $$\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}<2,$$and that finishes Given the title, you really had no reason to make this post... Another method: $$\frac{1}{n^2} < \int_{n-1}^n \frac{1}{x^2}\,dx$$$$S_n-1 < \int_1^n \frac{1}{x^2}\,dx = 1 - \frac{1}{n}$$$$\implies S_n < 2-1/n < 2$$
22.06.2021 05:09
Sorta equivalent to post 3 maybe, but \[S_n = 2 \left( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 4} + \frac{1}{3 \cdot 6} + \cdots \right) < 2 \left(\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \cdots \right)=2\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\cdots \right)=2\] edit: oops thanks @below
22.06.2021 05:20
, the $\le$ sign should be changed to $<$.
22.06.2021 06:35
Just use $k^2>k(k-1)$ and you get a telescoping sum.
22.06.2021 06:41
The.survivor wrote: Just use $k^2>k(k-1)$ and you get a telescoping sum. Wow, that's a good idea!
22.06.2021 07:51
The.survivor wrote: Just use $k^2>k(k-1)$ and you get a telescoping sum. Oh pfft what a way to avoid such a bash